
Calculate the mass percent of O present in sodium sulphate ( $ Na_2SO_4 $ ).
Answer
528.6k+ views
Hint :In the given question mass percent of oxygen is asked in a sodium sulphate fragment. Sodium sulphate compound consists of three elements sodium, Sulphur and oxygen combinedly known as sodium sulphate. We can calculate the mass percent of any element in the compound by using mass percent formula.
Complete Step By Step Answer:
Mass percent is the mass of the element or solute divided by the mass of the compound or solute. The result is multiplied by 100 to give a percent.
According to the given question
The molecular formula of sodium sulphate is $ \left( {N{a_2}S{O_4}} \right) $ .
To calculate mass percent, we have first found out the molar mass of an element.
Therefore, Molar mass = atomic mass of sodium + atomic mass of Sulphur + atomic mass of Oxygen
Or, we can write
Molar mass = atomic mass of Na (23) + atomic mass of S (32.066) + atomic mass of O (16)
So, Molar mass of $ \left( {N{a_2}S{O_4}} \right) $ is
$ \Rightarrow \left[ {\left( {2 \times 23.0} \right) + \left( {32.066} \right) + 4\left( {16.00} \right)} \right] = 142.066\;gmo{l^{ - 1}} $
The formula for the amount of mass percent of an element in a compound is:
$ Mass{\text{ }}percent{\text{ }}of{\text{ }}an{\text{ }}element\; = \dfrac{{mass{\text{ }}of{\text{ }}that{\text{ }}element{\text{ }}in{\text{ }}the{\text{ }}compound}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}compound}} \times 100 $
Molar mass of sodium sulphate is $ 142.066\;gmo{l^{ - 1}} $
Mass element of oxygen $ = 4 \times 16 = 64.0g $
By putting values in formula, we get;
Therefore, the mass percent of oxygen:
$ = \dfrac{{{\text{ }}64.0g}}{{142.66g}} \times 100\; $
$ = 45.049\% $
$ = 45.05\% $
Hence, the mass percentage of oxygen in sodium sulphate =45.05%.
Formula used: $ Mass{\text{ }}percent{\text{ }}of{\text{ }}an{\text{ }}element\; = \dfrac{{mass{\text{ }}of{\text{ }}that{\text{ }}element{\text{ }}in{\text{ }}the{\text{ }}compound}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}compound}} \times 100 $
Note :
Mass percentage is a way of presenting the concentration of an element in a compound or of a component within a mixture through which Mass percentage is calculated and multiplied by 100 and the mass of a component divided by the total mass of the mixture.
Complete Step By Step Answer:
Mass percent is the mass of the element or solute divided by the mass of the compound or solute. The result is multiplied by 100 to give a percent.
According to the given question
The molecular formula of sodium sulphate is $ \left( {N{a_2}S{O_4}} \right) $ .
To calculate mass percent, we have first found out the molar mass of an element.
Therefore, Molar mass = atomic mass of sodium + atomic mass of Sulphur + atomic mass of Oxygen
Or, we can write
Molar mass = atomic mass of Na (23) + atomic mass of S (32.066) + atomic mass of O (16)
So, Molar mass of $ \left( {N{a_2}S{O_4}} \right) $ is
$ \Rightarrow \left[ {\left( {2 \times 23.0} \right) + \left( {32.066} \right) + 4\left( {16.00} \right)} \right] = 142.066\;gmo{l^{ - 1}} $
The formula for the amount of mass percent of an element in a compound is:
$ Mass{\text{ }}percent{\text{ }}of{\text{ }}an{\text{ }}element\; = \dfrac{{mass{\text{ }}of{\text{ }}that{\text{ }}element{\text{ }}in{\text{ }}the{\text{ }}compound}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}compound}} \times 100 $
Molar mass of sodium sulphate is $ 142.066\;gmo{l^{ - 1}} $
Mass element of oxygen $ = 4 \times 16 = 64.0g $
By putting values in formula, we get;
Therefore, the mass percent of oxygen:
$ = \dfrac{{{\text{ }}64.0g}}{{142.66g}} \times 100\; $
$ = 45.049\% $
$ = 45.05\% $
Hence, the mass percentage of oxygen in sodium sulphate =45.05%.
Formula used: $ Mass{\text{ }}percent{\text{ }}of{\text{ }}an{\text{ }}element\; = \dfrac{{mass{\text{ }}of{\text{ }}that{\text{ }}element{\text{ }}in{\text{ }}the{\text{ }}compound}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}compound}} \times 100 $
Note :
Mass percentage is a way of presenting the concentration of an element in a compound or of a component within a mixture through which Mass percentage is calculated and multiplied by 100 and the mass of a component divided by the total mass of the mixture.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

